What are the solutions to the equation 2x^2+2=3x ?

The solutions to the equation 2x^2+2=3x are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4

Find the zeros of 2x^2+2=3x

The zeros of 2x^2+2=3x are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4

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2 x^{2} + 2 = 3 x

x = \frac{3}{4} + i \frac{7}{4}, x = \frac{3}{4} - i \frac{7}{4}

Solution steps

2 x^{2} + 2 = 3 x

Move

3 x

to the left side

2 x^{2} + 2 - 3 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - 3 x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

Simplify

(- 3)^{2} - 4 \cdot 2 \cdot 2 : 7 i

x_{1, 2} = \frac{- ( - 3 ) \pm 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 7 i}{2 \cdot 2}

x = \frac{- ( - 3 ) + 7 i}{2 \cdot 2} : \frac{3}{4} + i \frac{7}{4}

x = \frac{- ( - 3 ) - 7 i}{2 \cdot 2} : \frac{3}{4} - i \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{3}{4} + i \frac{7}{4}, x = \frac{3}{4} - i \frac{7}{4}

x^{2} + 5 x - 36

\frac{1}{1 - \frac{1}{4}}

∣ 10 - 4 x ∣ \leq 9

3 (y + 1)^{2} - 5 = 43

4 \cdot 1.5

What are the solutions to the equation 2x^2+2=3x ?
The solutions to the equation 2x^2+2=3x are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4
Find the zeros of 2x^2+2=3x
The zeros of 2x^2+2=3x are x= 3/4+i(sqrt(7))/4 ,x= 3/4-i(sqrt(7))/4